Question

State the principle of Potentiometre. Compare the electromotive force of two cells using potentiometre.​

Answer 1

• The potentiometer works on the principle that when a constant current flows through a wire of uniform cross sectional area, potential difference between its two points is directly proportional to the length of the wire between the two points.

• Comparison of EMFs of two given cells using potentiometer :

The potentiometer wire AB is connected in series with a battery (Bt), Key (K), rheostat (Rh) as shown in figure. This forms the primary circuit. The end A of potentiometer is connected to the terminal C of a DPDT switch (six-way key-double pole double throw). The terminal D is connected to the jockey (J) through a galvanometer (G) and high resistance (HR). The cell of emf E1  is connected between terminals C1   and D1  and the cell of emf E2   is connected between C2  and D2  of the DPDT switch.

Let I be the current flowing through the primary circuit and r be the resistance of the potentiometer wire per metre length.

The DPDT switch is pressed towards C1 , D1 so that cell E1 is included in the secondary circuit. The jockey is moved on the wire and adjusted for zero deflection in galvanometer. The balancing length is $I_{1}$ . The potential difference across the balancing length is $I_{1} = Irl_{1}$  .

Then, by the principle of potentiometer,

$E_{1} = iRL_{1}$            ....(1)

The DPDT switch is pressed towards $E_{2}$ .  The balancing length $I_{2}$ for zero deflection in galvanometer is determined. The potential difference across the balancing length is $I_{2} = Irl_{2}$

then  

$E_{2} = Irl_{2}$        ....(2)

Dividing (1) and (2) we get,

$\frac{E_{1}}{E_{2}} = \frac{I_{1}}{I_{2}}$

​If emf of one cell (E1 ) is known, the emf of the other cell (E2) can be calculated using the relation,

$E_{2} = E_{1} \frac{I_{1} }{I_{2}}$

​  

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Answer 2

Explanation:

The potentiometer works on the principle that when a constant current flows through a wire of uniform cross sectional area, potential difference between its two points is directly proportional to the length of the wire between the two points.

Comparison of EMFs of two given cells using potentiometer :

The potentiometer wire AB is connected in series with a battery (Bt), Key (K), rheostat (Rh) as shown in figure. This forms the primary circuit. The end A of potentiometer is connected to the terminal C of a DPDT switch (six-way key-double pole double throw). The terminal D is connected to the jockey (J) through a galvanometer (G) and high resistance (HR). The cell of emf E1 is connected between terminals C1 and D1 and the cell of emf E2 is connected between C2 and D2 of the DPDT switch.